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i'm preparing my calculus exam and i'm in doubt about how to generally compute triple integrals. I know that the cartesian equation of a sphere is $B_R=\{(x, y, z)|x^2+y^2+z^2=R^2\}$, so (if i didn't want to use spherical coordinates, wich i'm aware is the best way and i already did that) its volume would just be $\iiint_S dxdydz$, but what would the extremes be? I know $-R \leq z \leq R$ and $-\sqrt{R^2-y^2-z^2} \leq x \leq\sqrt{R^2-y^2-z^2} $, but what are the extremes for $y$? I can't describe it in terms of $x$, so i have $$Vol(B_R) = \int_{-R}^{R} \int_?^?\int_{-\sqrt{R^2-y^2-z^2}}^\sqrt{R^2-y^2-z^2} dxdydz.$$ What should be there instead of the '?'

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  • $\begingroup$ The limits are $\pm\sqrt{R^2-z^2}$, because that is the maximum/minimum that $y$ is allowed to take knowing $z$ only. $\endgroup$ – user10354138 Jun 8 at 15:51
  • $\begingroup$ So i have to compute the maximum/minimum of the function and substitute its $x$ coordinate in the extremes? $\endgroup$ – DottorMaelstrom Jun 8 at 15:55
  • $\begingroup$ I don't know what "function" you are talking about. Also, sphere is 2-dimensional, so doesn't have "volume". You want the ball $B_R=\{x^2+y^2+z^2\leq R^2\}$ instead. $\endgroup$ – user10354138 Jun 8 at 16:01
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After doing away with the integral over $x$ it remains to be integrated over $y$ and $z$, as you already know. The domain of the remaining double integral is a circle on the $YZ$ plane. So, its limits are $-\sqrt{R^2 - z^2}$ (lower) and $\sqrt{R^2 - z^2}$ (upper).

If you want to see it geometrically, think of your 2-dimensional sphere of radius $R$ in the 3-dimensional Cartesian space. Project everything onto the $YZ$ plane to get a flat 2-dimensional circle of radius $R$. You can see that circle as the union of infinitely many half-circumferences of radius $R$, for $R$ ranging from $0$ to $R$.

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  • $\begingroup$ So in general when computing multiple integrals, after being done with one variable, i should "forget it" and think the space as losing one dimension (the direction of the variable i already integrated)? $\endgroup$ – DottorMaelstrom Jun 8 at 16:11
  • $\begingroup$ Yes, but let me elaborate a bit on that. By integrating on a variable (say $x$) with those limits, you are passing the domain's information onto the integrand. When integrating a definite integral over $x$ your result cannot depend on $x$. With iterated integrals we follow this process with the hopes of obtaining a real number, which is the area or volume of a geometric object. $\endgroup$ – Sam Skywalker Jun 8 at 16:21
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It should be \begin{equation} \text{vol}(B_R) = \int_{-R}^{R} \int_{-\sqrt{R^2-z^2}}^{\sqrt{R^2-z^2}} \int_{-\sqrt{R^2-z^2-y^2}}^{\sqrt{R^2-z^2-y^2}} \, dx \, dy \, dz \end{equation} The way to think about this is to successively "fix" each variable as follows: pick a variable, for example $z$. Clearly, its bounds are $-R \leq z \leq R$. Then, for a fixed $z$, we have \begin{equation} 0 \leq x^2 + y^2 \leq R^2 - z^2. \end{equation} Now, pick your next variable, say $y$. From the above inequality, it follows that $y$ must satisfy $- \sqrt{R^2 - z^2} \leq y \leq \sqrt{R^2 - z^2}$ (if it was outside this interval, the inequality above would be violated). Now, keep $y$ fixed. Then, we get \begin{equation} 0 \leq x^2 \leq R^2 - z^2 - y^2 \end{equation} which implies $-\sqrt{R^2 - z^2 - y^2} \leq x \leq \sqrt{R^2 - z^2 - y^2}$.


So, what you're doing is you look at the constraint all $3$ variables satisfy. Then, you choose one of them (for example $z$) and ask yourself what are the bounds it can take (in this case it was obvious). Then, you think of $z$ as fixed, and ask yourself what constraint do the remaining two variables satisfy (in this case $x,y$). Then you pick one of them (for example $y$) and ask what values can it take. Now, think of $y$ as fixed ax well, so that you end up with a constraint on $x$. This is a step-by-step reduction process which should be helpful; to really solidify your understanding you should see what each step is doing geometrically as well.

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